Problem: Find the positive solution to
\[\sqrt[3]{x + \sqrt[3]{x + \sqrt[3]{x + \dotsb}}} = \sqrt[3]{x \sqrt[3]{x \sqrt[3]{x \dotsm}}}.\]
Solution: Let
\[y = \sqrt[3]{x \sqrt[3]{x \sqrt[3]{x \dotsm}}}.\]Then
\[y^3 = x \sqrt[3]{x \sqrt[3]{x \dotsm}} = xy,\]so $y^2 = x.$

Let
\[z = \sqrt[3]{x + \sqrt[3]{x + \sqrt[3]{x + \dotsb}}}.\]Then
\[z^3 = x + \sqrt[3]{x + \sqrt[3]{x + \dotsb}} = x + z,\]so $z^3 - z = x.$

Since $z = y,$ $y^3 - y = x = y^2.$  Then
\[y^3 - y^2 - y = 0,\]which factors as $y (y^2 - y - 1) = 0,$ so $y^2 - y - 1 = 0.$  By the quadratic formula,
\[y = \frac{1 \pm \sqrt{5}}{2}.\]Since $y$ is positive,
\[y = \frac{1 + \sqrt{5}}{2}.\]Then
\[x = y^2 = \boxed{\frac{3 + \sqrt{5}}{2}}.\]